Question:For the following equilibrium system, which of the changes will form more CaCO₃?

${\mathbf{CO}}_{\mathbf{2}}\left(g\right)\mathbf{+}\mathbf{Ca}{\left(\mathrm{OH}\right)}_{\mathbf{2}}\left(s\right)\mathbf{\to }{\mathbf{CaCO}}_{\mathbf{3}}\left(s\right)\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(l\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{∆}{\mathbf{H}}^{\mathbf{o}}\mathbf{=}\mathbf{-}\mathbf{113}\mathbf{kJ}$

(a) Decrease temperature at constant pressure (no phase change)

(b) Increase volume at constant temperature

(c) Increase partial pressure of CO₂

(d) Remove one-half of the initial CaCO₃

Given reaction is exothermic because$∆{\mathrm{H}}^{\mathrm{o}}$ is negative. If the enthalpy change is negative, then the reaction must be negative. There occurs a temperature decrease in heat releasing reactions at constant pressure. Exothermic reaction are forward reactions. So, decrease in temperature favors more yield of CaCO₃.

Decrease in pressure means there is an increase in volume. So, that reaction proceed left side to yield more gaseous molecules. Thus, increase in volume doesn’t favor more CaCo₃ in product side.

Partial pressures are mainly appeared in reaction quotient. They didn’t have any effect in reaction. So, addition of inert gas doesn’t affect the changes in partial pressure. So, the product side and reactant side remain same.

If a product concentration is decreased or it removed, then the reaction will shift towards backward direction. That means equilibrium shifts left side to reduce the disturbance occurred there. So, there is no chance of more product yield.